Derivative of (1/sqrt(x)+2)*tg(x)

1. Apply the quotient rule, which is:

   $$\frac{d}{d x} \frac{f{\left(x \right)}}{g{\left(x \right)}} = \frac{- f{\left(x \right)} \frac{d}{d x} g{\left(x \right)} + g{\left(x \right)} \frac{d}{d x} f{\left(x \right)}}{g^{2}{\left(x \right)}}$$

   $f{\left(x \right)} = \left(2 \sqrt{x} + 1\right) \tan{\left(x \right)}$ and $g{\left(x \right)} = \sqrt{x}$.

   To find $\frac{d}{d x} f{\left(x \right)}$:

   1. Apply the product rule:

      $$\frac{d}{d x} f{\left(x \right)} g{\left(x \right)} = f{\left(x \right)} \frac{d}{d x} g{\left(x \right)} + g{\left(x \right)} \frac{d}{d x} f{\left(x \right)}$$

      $f{\left(x \right)} = 2 \sqrt{x} + 1$; to find $\frac{d}{d x} f{\left(x \right)}$:

      1. Differentiate $2 \sqrt{x} + 1$ term by term:

         1. The derivative of the constant $1$ is zero.

         2. The derivative of a constant times a function is the constant times the derivative of the function.

            1. Apply the power rule: $\sqrt{x}$ goes to $\frac{1}{2 \sqrt{x}}$

            So, the result is: $\frac{1}{\sqrt{x}}$

         The result is: $\frac{1}{\sqrt{x}}$

      $g{\left(x \right)} = \tan{\left(x \right)}$; to find $\frac{d}{d x} g{\left(x \right)}$:

      1. Rewrite the function to be differentiated:

         $$\tan{\left(x \right)} = \frac{\sin{\left(x \right)}}{\cos{\left(x \right)}}$$

      2. Apply the quotient rule, which is:

         $$\frac{d}{d x} \frac{f{\left(x \right)}}{g{\left(x \right)}} = \frac{- f{\left(x \right)} \frac{d}{d x} g{\left(x \right)} + g{\left(x \right)} \frac{d}{d x} f{\left(x \right)}}{g^{2}{\left(x \right)}}$$

         $f{\left(x \right)} = \sin{\left(x \right)}$ and $g{\left(x \right)} = \cos{\left(x \right)}$.

         To find $\frac{d}{d x} f{\left(x \right)}$:

         1. The derivative of sine is cosine:

            $$\frac{d}{d x} \sin{\left(x \right)} = \cos{\left(x \right)}$$

         To find $\frac{d}{d x} g{\left(x \right)}$:

         1. The derivative of cosine is negative sine:

            $$\frac{d}{d x} \cos{\left(x \right)} = - \sin{\left(x \right)}$$

         Now plug in to the quotient rule:

         $\frac{\sin^{2}{\left(x \right)} + \cos^{2}{\left(x \right)}}{\cos^{2}{\left(x \right)}}$

      The result is: $\frac{\left(2 \sqrt{x} + 1\right) \left(\sin^{2}{\left(x \right)} + \cos^{2}{\left(x \right)}\right)}{\cos^{2}{\left(x \right)}} + \frac{\tan{\left(x \right)}}{\sqrt{x}}$

   To find $\frac{d}{d x} g{\left(x \right)}$:

   1. Apply the power rule: $\sqrt{x}$ goes to $\frac{1}{2 \sqrt{x}}$

   Now plug in to the quotient rule:

   $\frac{\sqrt{x} \left(\frac{\left(2 \sqrt{x} + 1\right) \left(\sin^{2}{\left(x \right)} + \cos^{2}{\left(x \right)}\right)}{\cos^{2}{\left(x \right)}} + \frac{\tan{\left(x \right)}}{\sqrt{x}}\right) - \frac{\left(2 \sqrt{x} + 1\right) \tan{\left(x \right)}}{2 \sqrt{x}}}{x}$

2. Now simplify:

   $$\frac{2 x^{\frac{3}{2}} + x - \frac{\sin{\left(2 x \right)}}{4}}{x^{\frac{3}{2}} \cos^{2}{\left(x \right)}}$$

The answer is:

$$\frac{2 x^{\frac{3}{2}} + x - \frac{\sin{\left(2 x \right)}}{4}}{x^{\frac{3}{2}} \cos^{2}{\left(x \right)}}$$